\section{Automorphisms and Galois Groups \cite{book:Fraleigh2003}} 

Automorphisms are a special type of isomorphism. It it sensible therefore to start the discussion with some generalities on isomorphism which will be used in the study of automorphisms. Automorphisms appear in the topic of coding theory when it comes to skew polynomial rings.

\begin{Def} [Conjugate]
Two elements $a,b \in E$, an algebraic extension of $F$, are {\bf conjugate} over $F$ if they are zeros to the same irreducible polynomial over $F$.
\end{Def}

\begin{Thrm} [The Conjugation Isomorphism] \label{Thrm:ConjIso}
Let $F$ be a field and $\alpha$ and $\beta$ two algebraic elements over $F$. The mapping $\theta: F(\alpha) \rightarrow F(\beta)$ defined by
\[
\theta_{\alpha,\beta}(c_{0}+c_{1}\alpha+ \dots +c_{n-1}\alpha^{n-1}) \rightarrow c_{0}+c_{1}\beta+ \dots +c_{n-1}\beta^{n-1}
\]
is an isomorphism if and only if $\alpha$ and $\beta$ are conjugate, where $c_{0}, \dots , c_{n-1} \in F$.
\end{Thrm}

Theorem \ref{Thrm:ConjIso} shows that to define an isomorphism from one extension field to another that faithfully maps every element of the field to itself, the algebraic elements must map to their conjugates. The following terminology is useful in dealing with such cases.

\begin{Def} [Left Fixed]
An element $a$ that maps to itself under some isomorphism $\theta$ is said to be {\bf left fixed} under $\theta$. That is, $\theta$ leaves $a$ fixed. 
\end{Def}

\begin{eg} \label{eg:ConjIso1}
Consider the algebraic extensions $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(-\sqrt{2})$ of $\mathbb{Q}$. Define an isomorphism $\theta: \mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(-\sqrt{2})$ with:
\[
\theta(\sqrt{2})=-\sqrt{2}
\]
and
\[
\theta(\alpha)=\alpha, \alpha \in \mathbb{Q}
\]
Then $\theta$ is an isomorphism leaving $\mathbb{Q}$ fixed and mapping $\sqrt{2}$ to $-\sqrt{2}$. The notation $\theta_{\sqrt{2},-\sqrt{2}}$ is some times used to show this result.
\end{eg}

\begin{eg}
Let $f(x) \in \mathbb{R}[x]$ and $a+ib \in \mathbb{C}$ such that $f(a+ib)=0$. The claim is that $f(a-ib)=0$ also. Let's consider the polynomial $x^2+1 \in \mathbb{R}[x]$ which leads to $\mathbb{C}=\mathbb{R}(i)$. In this case it is easy to see that both $i$ and $-i$ are zeros of $x^{2}+1$ and are conjugate. Following the notation in Example \ref{eg:ConjIso1}, we can then define an isomorphism $\theta_{i,-i}$. Let:
\[
f(x) = c_{0}+c_{1}x+ \dots + c_{n-1}x^{n-1}, c_{0}, \dots, c_{n-1} \in \mathbb{R}
\] 
We have:
\[
f(a+ib) = c_{0}+c_{1}(a+ib)+ \dots + c_{n-1}(a+ib)^{n-1} = 0
\]
Applying the isomorphism $\theta_{i,-i}$ defined above:
\begin{eqnarray}
0 = \theta(0) &=& \theta_{i,-i}(f(a+ib)) 	= \theta_{i,-i}(c_{0}+c_{1}(a+ib)+ \dots + c_{n-1}(a+ib)^{n-1})	\nonumber \\
			&=& \theta_{i,-i}(c_{0})+\theta_{i,-i}(c_{1}(a+ib))+ \dots + \theta_{i,-i}(c_{n-1}(a+ib)^{n-1}) \nonumber \\
			&=& \theta_{i,-i}(c_{0})+\theta_{i,-i}(c_{1})\theta_{i,-i}(a+ib)+ \dots + \theta_{i,-i}(c_{n-1})\theta_{i,-i}^{n-1}(a+ib) \nonumber \\
			&=& c_{0} + c_{1}\theta_{i,-i}(a+ib)+ \dots + c_{n-1}\theta_{i,-i}^{n-1}(a+ib) \nonumber \\
			&=& c_{0} + c_{1}(a-ib)+ \dots + c_{n-1}(a-ib)^{n-1} \nonumber \\
			&=& f(a-ib) \nonumber
\end{eqnarray}
\end{eg}

\begin{Def} [Automorphism]
An isomorphism $\theta: F \rightarrow F$ from a field $F$ to itself is called an {\bf automorphism}.
\end{Def}

By this definition, it is clear that automorphism is merely a special type of isomorphism, preserving addition and multiplication, while in this case the domain and co-domain are the same. As such, all the discussion above equally applies to automorphisms. In what follows those concepts are carried on but more specific to automorphisms.

\begin{Thrm}
The set of all the automorphisms of a field $E$ forms a group under composition called the {\bf group of automorphisms} of $E$.
\end{Thrm}

\begin{Thrm}
Let $\{\sigma_{i} | i \in I\}$ be a set of automorphisms over a field $E$. The set of elements in $E$ are left fixed under $\sigma_{i}$'s is a subfield of $E$.
\end{Thrm}

\begin{Thrm} [Galois Group]
Let $F$ be a subfield of $E$. The set of automorphisms over $E$ that leave $F$ fixed is a subgroup of the group of automorphisms of $E$ called the {Galois Group} of $E$ and denoted by $Gal(E/F)$. 
\end{Thrm}

\begin{Thrm} [Frobenius Automorphism] \label{Thrm:Frobenius}
Let $F$ be a finite field of characteristic $p$. Then the map $\sigma_{p}: F \rightarrow F$ defined by $\sigma_{p}(a)=a^{p}$ for $a \in F$ is an automorphism called the {\bf Frobenius Automorphism} of $F$. Furthermore, the subfield of $F$ left fixed under $\sigma_{p}$ is isomorphic to $\mathbb{Z}_{p}$.  
\end{Thrm}
\begin{proof}
Let $a,b \in F$. We have:
\[
\sigma_{p}(ab)=(ab)^{p}=a^{p}b^{p}=\sigma_{p}(a)\sigma_{p}(b)
\]
and,
\[
\sigma_{p}(a+b)=(a+b)^{p}=a^{p}+b^{p}=\sigma_{p}(a)+\sigma_{p}(b)
\]
since $\text{Char}(F)=p$. The kernel of $\sigma_{p}$ is $\{0\}$ since:
\[
\sigma_{p}(a)=a^{p}=0 \Rightarrow a=0
\]
Therefore $\sigma_{p}$ is an automorphism. The group of elements $x$ that are left fixed under $\sigma_{p}$ satisfy:
\[
x^{p}=x \Rightarrow x^{p}-x=0
\]
The zeros of the equation on the right hand side are the elements of $\mathbb{Z}_{p}$.
\end{proof}

\begin{Thrm} [Galois Groups over Finite Fields] \label{Thrm:FiniteAuto}
Let $E$ be a finite extension of degree $n$ of a field $F$ of $p^{r}$ elements. Then the Galois Group $Gal(E/F)$ is cyclic of order $n$, and is generated by $\sigma_{p^{r}}$, the Frobenius Automorphism, where $\sigma_{p^{r}}(\alpha)=\alpha^{p^{r}}$ for every $\alpha \in E$.
\end{Thrm}

Theorem \ref{Thrm:FiniteAuto} shows that the group of all automorphisms of a finite field is cyclic and has a canonical (natural) generator, namely the \emph{Frobenius Automorphism}. The following example brings together the concepts discussed above and has extensive application in the studies of coding theory.

\begin{eg} [*important]
Let us consider $F_{4}=\{0,1,\alpha,\alpha^{2}=1+\alpha\}$ an extension of $\mathbb{Z}_{2}$ of dimension $2$ given by $F_{2}[x]/<x^{2}+x+1>$, where the primitive element of $F_{4}$ is denoted by $\alpha$. By Theorem \ref{Thrm:FiniteAuto} we know $G(F_{4}/\mathbb{Z}_{2})$ has order $2$ and is generated by $\sigma_{2}$. Let us verify. 

It can be observed that:
\[
\sigma_{2}(0)=0^{2}=0 \text{ and } \sigma_{2}(1)=1^{1}=1.
\]     
Thus the $\{0,1\}$ is left fixed which is no coincidence by Theorem \ref{Thrm:Frobenius} we have $\mathbb{Z}_{2}=\{0,1\}$. Next consider:
\[
\sigma_{2}(\alpha)=\alpha^{2}
\]
Therefore $\alpha$ is mapped to $\alpha^{2}$ which is another element in $F_{4}$. By \ref{Thrm:ConjIso}, since $\sigma_{2}$ is an automorphism, hence an isomorphism, then $\alpha$ and $\alpha^{2}$ must be conjugate. This is in fact true because:
\[
x^{2}+x+1=(x-\alpha)(x-\alpha^{2})
\]
\end{eg}

 






